It comes down to whether the centre has a faithful state or not, and an abelian von Neumann algebra has a faithful state iff it is $\sigma$-finite, i.e. iff every disjoint family of non-zero projections is countable. A finite factor is of course tracial because its centre is isomorphic to $\mathbb{C}$.

Archived here, Thom's remark starts at 14:46

Compact Hausdorff spaces also have a "minimal" cover by a Stonean space, which is obtained by taking the Stone space of the complete Boolean algebra of regular open sets (a construction due to Andrew Gleason).

$\mathbf{Prof}^{\mathrm{op}}$ is equivalent to the category of Boolean algebras and therefore locally $\aleph_0$-presentable. The category $\mathbf{CHaus}^{\mathrm{op}}$ is not locally $\aleph_0$-presentable, but it is locally $\aleph_1$-presentable (and in fact monadic over $\mathbf{Set}$) because it is equivalent to the category of commutative unital C$^*$-algebras. This can be stated as $\mathbf{CHaus}$ is the free completion of the compact metric spaces (equivalently, the closed subspaces of $[0,1]^\mathbb{N}$). Is this the kind of thing you're looking for?

You can build the algebra as a quotient of the free product $\mathbb{C}^2 \ast \mathbb{C}^2 \ast C([-1,1])$, where $x$ and $y$ are $(1,0)$ interpreted in the first and second algebras and $z$ is the function $\alpha \mapsto i\alpha : [-1,1] \rightarrow \mathbb{C}$ in the third algebra. You take a quotient to impose your relations $[x,z] = y$ and $[y,z] = -x$.

Carlo Beenakker's suggestion works in any W$^*$-algebra by the Borel functional calculus for normal elements, but interestingly it is not the case that for any C$^*$-algebra $A$, every unitary $u \in A$ is of the form $e^{ih}$ for a self-adjoint $h$. Consider the C$^*$-algebra $C(S^1)$, where $S^1$ is the unit complex numbers. For a self-adjoint $h \in C(S^1)$, the function $t \mapsto e^{i(1-t)h} : [0,1] \to C(S^1)$ defines a homotopy from $e^{ih}$ to $1$, so $e^{ih}$ has winding number zero. Therefore the unitary $z : S^1 \hookrightarrow \mathbb{C}$ is not $e^{ih}$ for any self-adjoint $h$.

@JamieGabe Nice argument! Please post this as an answer so the question can be marked as solved.

@BenjaminSteinberg I will summarize my comments into an answer, but I'm not arguing "slightly backwards", I'm giving an argument that only assumes consistency of ZFC rather than soundness.

@BenjaminSteinberg We really use the halting problem, not just computational undecidability. There are undecidable problems that are weaker than the halting problem, and the proof wouldn't work with those.

@BenjaminSteinberg I thought someone might say that. ZFC has a recursively enumerable set of axioms, so we can make a Turing machine $T$ that enumerates all formal proofs starting from the ZFC axioms and halts if it reaches a proof of $0 = 1$. Then $\mathrm{Con}(\mathrm{ZFC})$ is equivalent to the halting problem for $T$.

Does "are these two 4-manifolds diffeomorphic" count as differential geometry? Andrei Markov Jr showed how to take two finite group presentations and build two 4-manifolds that are diffeomorphic iff the corresponding groups are isomorphic. It is possible to construct a finite presentation that presents the trivial group iff $\mathrm{Con}(\mathrm{ZFC})$ is true. By Gödel's second incompleteness theorem, it is therefore independent of ZFC if the two 4-manifolds are diffeomorphic.

It's actually "Topological Psychology", not "Psychological Topology". It is not intended to be a joke, but it's certainly good for a chuckle.